Two parallel-plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates, while Y contains a dielectric medium of *ε _{r}* = 4.

(i) Calculate capacitance of each capacitor if the equivalent capacitance of the combination is 4 μF.

(ii) Calculate the potential difference between the plates X and Y.

(iii) Estimate the ratio of electrostatic energies stored in X and Y.

#### Solution

(i) We know that the capacitance of a parallel-plate capacitor is given by

`C=(varepsilon_0varepsilon_rA)/d`

`C_Y=(varepsilon_0 4A)/d`

`C_X=(varepsilon_0 A)/d`

Thus,

`=>C_Y=4C_X`

`=>C_(eq)=(C_xC_Y)/(C_x+C_Y) = 4muF`

`=>(C_X(4C_X))/(5C_X)=4muF`

`=>C_X=5muF`

`=>C_Y=20muF`

(ii) The potential difference between plates X and Y can be calculated as follows:

Q=CV

`=>C_XV_X=C_YV_Y`

`=>V_X/V_Y=C_X/C_Y=4`

`=>V_X=4V_Y`

Also, V_{Y}+V_{X}=15

⇒V_{Y}=3 V

⇒V_{X}=12 V

(iii) The ratio of electrostatic energies can be calculated as follows:

`E=(Q^2)/(2C)`

`=>E_X/E_Y=C_Y/C_X=4`

`=>E_X/E_Y=4/1`